Trigo problem?

Question

At a certain instance the captain of the ship observed that the angle of elevation of the top of the lighthouse is 50 meters above the water level is 60 degrees. After travelling directly away from the lighthouse for 5minutes, the same captain noticed that the angle of elevation is now 30 degrees. If the telescope is 6 meters above the waterline, find the velocity of the ship

Answer 1

You will be able to draw 2 right triangles.

The 1st right triangle represents the instance when the angle of elevation from the telescope to the top of the lighthouse was 60 degrees.
The height for this triangle is 44 ( 50 - 6 ) meters
The base is unknown yet, lets call it "a"
The angle opposite side 44 measures 60 degrees

Therefore tan 60 = 44/a
but we know that tan 60 is also = sqrt 3
hence 44/a = sqrt 3
a = 44/sqrt 3

The 2nd triangle represents the instance when the angle of elevation was 30 degrees
The height for this triangle is still 44 meters
But the base is now longer, it is = a + the distance travelled by the ship in 5 minutes
Lets call this distance d, therefore the base = a + d = (44/sqrt3) + d
the angle opposite side 44 is now 30 degrees

Therefore tan 30 = 44/[(44/sqrt 3) + d]
but we know that tan 30 = 1/sqrt 3
hence 1/ sqrt 3 = 44/[(44/sqrt 3) + d]
Simplifying, d = 88/sqrt3 meters

Speed = distance/time
Speed = (88/sqrt 3) meters / 5 minutes
Speed = 88/(5sqrt3) meters/minute
Rationalizing,
Speed = (88 sqrt 3) /15 meters/minute

Answer 2

The problem is worded clumsily. Is the lighthouse 50m tall or is the line of site from the telescope to the top of the lighthouse 50m/ Assuming the lighthouse is 50m tall:

d1 = 44/tan60 = 25
d1 = 44/tan30 = 76

Speed = (76-25)/5 = 10 m/min

All answers are to 2 significant figures. Any further precision is meaningless.

Answer 3

at the first point the ship is 25.4 meters away
the second point the ship is 76.2 meters away
so the ship traveled 50.8 metersin five minutes or 10.16 meters per minute.